If f(x)= ${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ then $\underset{x\to \frac{1}{2}}{\lim }\frac{2\left[f\left(x\right)-f\left({\displaystyle \frac{1}{2}}\right)\right]}{2x-1}=$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1}\left(\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{1-\sqrt{{\displaystyle \frac{1-\mathrm{x}}{1+\mathrm{x}}}}}{1+\sqrt{{\displaystyle \frac{1-\mathrm{x}}{1+\mathrm{x}}}}}\right) \\ \mathrm{x}=\cos 2\mathrm{\theta }=\mathrm{\theta }=\frac{1}{2}{\cos }^{-1}\mathrm{x} \\ ={\tan }^{-1}\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right) \\ =\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\cos }^{-1}\mathrm{x} \\ \underset{\mathrm{x}\to \frac{1}{2}}{\mathrm{Lt}}\frac{2\left[\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left({\displaystyle \frac{1}{2}}\right)\right]}{2(\mathrm{x}-{\displaystyle \frac{1}{2}})} \\ ={\mathrm{f}}^1\left(\frac{1}{2}\right) \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=0-\frac{1}{2}\left(\frac{-1}{\sqrt{1-{\mathrm{x}}^2}}\right)=\frac{1}{2\sqrt{1-{\mathrm{x}}^2}} \\ {\mathrm{f}}^1\left(\frac{1}{2}\right)=\frac{1}{\sqrt{3}} \end{gathered}$$